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Chapter 6: Problem 37

\(\mathrm{A} 6.0 \mathrm{~kg}\) box moving at \(3.0 \mathrm{~m} / \mathrm{s}\) on a horizontal, frictionless surface runs into one end of a light horizontal spring of force constant \(75 \mathrm{~N} / \mathrm{cm}\) that is fixed at the other end. Use the work-energy theorem to find the maximum compression of the spring.

Short Answer

Expert verified
Applying the known values \( m = 6.0 \, \mathrm{kg} \), \( v_i = 3.0 \, \mathrm{m/s} \), and \( k = 7500\, \mathrm{N/m} \) into the formula, \( x = \sqrt{\frac{mv_i^2}{k}} \), gives \( x \approx 0.14 \, \mathrm{m} \) or 14 cm as the maximum compression of the spring.

Step by step solution

01

Identify known variables.

We know the mass of the box \( m = 6.0 \, \mathrm{kg} \), the initial velocity of the box \( v_i = 3.0 \, \mathrm{m/s} \), and the spring constant \( k = 75 \, \mathrm{N/cm} = 7500\, \mathrm{N/m} \). Since the box eventually comes to rest due to the spring force, the final velocity \( v_f = 0 \, \mathrm{m/s} \). We have to find the maximum compression \( x \) of the spring.
02

Use the work-energy theorem.

The work-energy theorem implies that the work done on an object is equal the change in its kinetic energy. Since all the kinetic energy of the box is being transferred to the spring, we have the work done \( W = \frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2 \) which simplifies to \( W = \frac{1}{2}mv_i^2 \) as \( v_f = 0 \).
03

Apply the formula for spring potential energy.

The work done by the box on the spring is stored as potential energy \( U = \frac{1}{2}kx^2 \). We want this potential energy to equal the work done, so we set \( \frac{1}{2}kx^2 = \frac{1}{2}mv_i^2 \).
04

Solve for the spring compression.

From \( \frac{1}{2}kx^2 = \frac{1}{2}mv_i^2 \), cancel out the \(\frac{1}{2}\)'s and isolate \( x \), the maximum compression of the spring. Hence, \( x = \sqrt{\frac{mv_i^2}{k}} \). Substituting the known values allows us to calculate the spring compression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Potential Energy
When we talk about spring potential energy, we are referring to the energy stored in a spring when it is compressed or stretched. This is a type of potential energy because it has the potential to do work when the spring is released back to its relaxed state. For a spring, this stored energy depends on how much the spring is compressed or stretched and on the spring's stiffness.

The formula used to calculate this energy is:
  • \( U = \frac{1}{2}kx^2 \)
where:
  • \( U \) is the spring potential energy,
  • \( k \) is the spring constant, representing how stiff the spring is, and
  • \( x \) is the displacement from the spring's equilibrium position.
This formula shows us that the potential energy increases with both a higher spring constant and a greater displacement. Understanding spring potential energy is crucial when analyzing motion problems involving springs, as it helps us predict the system's behavior when forces are applied.
Kinetic Energy
Kinetic energy is the energy of motion. Anything that moves has kinetic energy, and this energy depends on the mass of the object and its velocity. When we talk about the kinetic energy of an object, we're basically discussing how much energy it has due to its motion.

The formula for kinetic energy is:
  • \( KE = \frac{1}{2}mv^2 \)
where:
  • \( KE \) is the kinetic energy,
  • \( m \) is the mass of the object, and
  • \( v \) is the velocity of the object.
As you can see, the kinetic energy increases with the square of the velocity. This means that an object moving twice as fast will have four times the kinetic energy. In our exercise, when the box compresses the spring, the kinetic energy is transformed into spring potential energy. This transformation is a great example of energy conservation in action!
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It tells us how resistant a spring is to being compressed or stretched. Simply put, a spring with a higher spring constant is stiffer than one with a lower spring constant.

In the context of the exercise, the spring constant helps us calculate how much energy the spring can store when it is compressed by the moving box. It is essential when working with Hooke's Law, which is the foundation for understanding how springs behave under force.
  • The formula in terms of force is \( F = kx \), where \( F \) is the force applied to the spring and \( x \) is the displacement.
Thus, the spring constant is critical in determining not only the stored energy via spring potential energy but also how effectively the spring performs when responding to external forces. Knowing a spring's constant allows us to predict how a spring will behave in dynamic situations, like in our exercise, where the spring is being compressed by a sliding box.

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Most popular questions from this chapter

A boxed \(10.0 \mathrm{~kg}\) computer monitor is dragged by friction \(5.50 \mathrm{~m}\) upward along a conveyor belt inclined at an angle of \(36.9^{\circ}\) above the horizontal. If the monitor's speed is a constant \(2.10 \mathrm{~cm} / \mathrm{s}\), how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

A force \(\vec{F}\) that is at an angle \(60^{\circ}\) above the horizontal is applied to a box that moves on a horizontal frictionless surface, and the force does work \(W\) as the box moves a distance \(d\). (a) At what angle above the horizontal would the force have to be directed in order for twice the work to be done for the same displacement of the box? (b) If the angle is kept at \(60^{\circ}\) and the box is initially at rest, by what factor would \(F\) have to be increased to double the final speed of the box after moving distance \(d ?\)

Using a cable with a tension of \(1350 \mathrm{~N}\), a tow truck pulls a car \(5.00 \mathrm{~km}\) along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

A 4.80 kg watermelon is dropped from rest from the roof of an 18.0 m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what are the watermelon鈥檚 (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

A block of ice with mass \(2.00 \mathrm{~kg}\) slides \(1.35 \mathrm{~m}\) down an inclined plane that slopes downward at an angle of \(36.9^{\circ}\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.
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